If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles.

OO' is the line segment joining the centers.
Let OO' intersect AB at M
Now Draw line segments OA, OB , O'A and O'B
In ΔOAO' and OBO' , we have
OA = OB (radii of same circle)
O'A = O'B (radii of same circle)
O'O = OO' (common side)
⇒ ΔOAO' ≅ ΔOBO' (SSS congruency)
⇒ ∠AOO' = ∠BOO'
⇒ ∠AOM = ∠BOM ......(i)
Now in ΔAOM and ΔBOM we have
OA = OB (radii of same circle)
∠AOM = ∠BOM (from (i))
OM = OM (common side)
⇒ ΔAOM ≅ ΔBOM (SAS congruency)
⇒ AM = BM and ∠AMO = ∠BMO
But
∠AMO + ∠BMO = 180°
⇒ 2∠AMO = 180°
⇒ ∠AMO = 90°
Thus, AM = BM and ∠AMO = ∠BMO = 90°
Hence OO' is the perpendicular bisector of AB.